This is my first post here. I apologize if it goes against any guidelines for posting. I study math as a hobby and am currently dealing with trigonometry on a high school level. I have so far learned the formulas for trigonometric addition and subtraction and double the angle, as well as what is in my language referred to as the ’trigonometric one’ - getting the radius of the unit circle by use of the pythagorean theorem. I have not yet gotten to deriving trigonometric functions. The following is a problem I could solve by plugging in a set of numbers, but in seeking a more elegant solution, perhaps, I found myself stuck and I don’t know what I am missing. I am appreciative of any help I get. The problem is as follows:

Show that if $A$ is an angle and $0^\circ<A<90^\circ$ then $\hspace{0.3cm}\left( 1+\dfrac {1}{\sin A}\right) \left( 1+\dfrac {1}{\cos A}\right)>5$

I began with the following assumption:

$$0^\circ<A<90^\circ\rightarrow0<{\sin A}<1\\0<{\cos A}<1\rightarrow\dfrac {1}{\sin A}\\\dfrac {1}{\cos A}>1$$

Given the above, it would follow that:

$$\begin{aligned} \lim _{A\rightarrow 90^\circ}\dfrac {1}{\cos A}&=\infty \\ \lim _{A\rightarrow 0^\circ}\dfrac {1}{\sin A}&=\infty \end{aligned}$$

This alone doesn’t seem like enough to show what is asked. I can show that at $A=45^\circ$ the product is still greater than 5, but I am not sure how any offset in degrees from there affects two trigonometric terms such that the product is still greater than 5. I also tried solving the inequality but ended up with fractioned terms I couldn’t add up or a cubic function if you will, that I couldn’t solve.

  • 1
    $\begingroup$ do you know trig identities such as $\sin^2 x+\cos^2 x=1$? $\endgroup$ – Vasya Jul 2 '18 at 15:42
  • 2
    $\begingroup$ Ignore $\lim_{A\to 90}$ We care about all values not just the the ones on the edges. $\endgroup$ – fleablood Jul 2 '18 at 15:48
  • 1
    $\begingroup$ You shouldn't use $\to$ to mean $\implies.$ They denote entirely different concepts. $\endgroup$ – Allawonder Jul 2 '18 at 22:08
  • $\begingroup$ Vasya, yes I do. I just didn’t know if it had a particular name. $\endgroup$ – Grenadine Jul 4 '18 at 15:50


Since $A$ is sharp you can write $\sin A = b/c$ and $\cos A = a/c$ where $c$ is hypotenuse in right triangle $ABC$ ($C=90$). Does that help?

Any way, since for all positive $x,y$ we have $x+y\geq 2\sqrt{xy}$: $$1+{1\over \sin A}\geq {2\over \sqrt{\sin A}}$$ and $$1+{1\over \cos A}\geq {2\over \sqrt{\cos A}}$$

so $$(1+{1\over \sin A})(1+{1\over \cos A})\geq {4\sqrt{2}\over \sqrt{\sin 2A}}\geq 4\sqrt{2} >5$$

  • $\begingroup$ Thank you Angle! $\endgroup$ – Grenadine Jul 4 '18 at 15:52

They probably expected something like this:

Expanding out, your inequality is the same as $$1 + {1 \over \sin A} + {1 \over \cos A} + {1 \over \sin A \cos A} \geq 5$$ This is equivalent to $${1 \over \sin A} + {1 \over \cos A} + {1 \over \sin A \cos A} \geq 4$$ Which is the same as $${1 \over \sin A} + {1 \over \cos A} + {2 \over \sin 2A} \geq 4$$ Since in the range in question, $0 < \sin A, \cos A < 1$ and $0 < \sin 2A \leq 1$, one has $${1 \over \sin A} + {1 \over \cos A} + {2 \over \sin 2A} > 1 + 1 + 2$$ $$= 4$$


If we open the parenthesis, we get $$1+{1\over \sin A}+ {1\over \cos A}+{1\over \sin A \cos A}\geq 5$$ $${1+\sin A+ \cos A\over \sin A \cos A}\geq 4$$ $$1+\sin A+ \cos A -4 \sin A \cos A \ge 0$$ $$(\cos A - \sin A)^2+\cos A(1-\sin A)+\sin A(1-\cos A) \ge 0$$ In the last inequality, each term is not negative so the sum is not negative


@Angle has given a simple way of solving the problem. If you know a little calculus, you could do it this way.

Let $$f(A)=\left( 1+\dfrac {1}{\sin A}\right) \left( 1+\dfrac {1}{\cos A}\right).$$ Then by the Quotient Rule, $$\small f'(A)=\left(0+\frac{0\sin A-1\cos A}{\sin^2A}\right)\left( 1+\dfrac {1}{\cos A}\right)+\left( 1+\dfrac {1}{\sin A}\right)\left( 0+\frac{0\cos A-1(-\sin A)}{\cos^2A}\right)$$ so $$\small f'(A)=-\frac{\cos A}{\sin^2A}\left( 1+\dfrac {1}{\cos A}\right)+\frac{\sin A}{\cos^2A}\left( 1+\dfrac {1}{\sin A}\right)=-\frac{\cos A}{\sin^2A}-\frac1{\sin^2A}+\frac{\sin A}{\cos^2A}+\frac1{\cos^2A}$$ giving $$f'(A)=\frac{1+\sin A}{\cos^2A}-\frac{1+\cos A}{\sin^2A}=0$$ for stationary points.

Thus $$\sin^2A(1+\sin A)=\cos^2A(1+\cos A)\implies \sin^2A-\cos^2A+\sin^3A-\cos^3A=0$$ so $$(\sin A-\cos A)(\sin A+\cos A)+(\sin A-\cos A)(\sin^2A+\sin A\cos A+\cos^2A)=0$$ giving $$(\sin A-\cos A)(1+\sin A+\cos A+\sin A\cos A)=0$$ and clearly one solution is when $\tan A=1\implies A=45^\circ$.

For the other equation, we solve $$1+\sin A+\cos A+\sin A\cos A=(1+\sin A)(1+\cos A)=0$$ but the solutions are outside of the range of $A$.

Hence $A=45^\circ$.

Now at this angle, $$f(45^\circ)=(1+\sqrt2)^2=3+2\sqrt2>3+2=5.$$

  • 1
    $\begingroup$ How do you deduce that $\sin^2 A + \sin^3 A = \cos^2 A + \cos^3 A$ implies that $\sin^2 A = \cos^2 A$ and $\sin^3 A = \cos^3 A$? When the latter two equalities are satisfied, the first clearly is, so we have a stationary point; but I don't see why (in this argument) there can be no other stationary points, which might have smaller values. $\endgroup$ – tomsmeding Jul 2 '18 at 20:23
  • $\begingroup$ @tomsmeding I agree. Please see the edit now. $\endgroup$ – TheSimpliFire Jul 3 '18 at 5:49
  • $\begingroup$ That looks great! $\endgroup$ – tomsmeding Jul 3 '18 at 7:58
  • $\begingroup$ Thanks, TheSimpleFire. I know some calculus but I haven’t gotten to things like the quotient rule yet. $\endgroup$ – Grenadine Jul 4 '18 at 15:49

Use $AM-GM$ and $\sin^2 A + \cos^2 A = 1$.


I start by simply doing.

$(1 + \frac {1}{\sin A})(1 + \frac 1{\cos A}) = $

$(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A}$.

Now $0 < \sin A < 1$ so $ \frac 1{\sin A} > 1$ and $0 < \cos A < 1$ so $ \frac 1{\cos A} > 1$.

So $(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A} > 3 + + \frac 1{\sin A \cos B}$.

Now common sense tells us $\sin A \cos A < 1*1$ so $\frac 1{\sin A \cos A} > 1$ and that gives us:

$(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A}) > 4$ and that is not good enough.

So I need more. Now I know $\sin A$ and $\cos A$ aren't just any numbers less than one. I know $\sin^2 A + \cos^2 A = 1$ and I know that if gets close to $1$ or $0$ the other gets close to $0$ and $1$ and I know that $\sin 45 = \cos 45 = \frac{\sqrt {2}}{2}$ but other wise one is more and the other is less than $\frac{\sqrt{2}}{2}$.

This all triggers that I can probably use the AM-GM theorem to "boost" $\frac 1 {\sin A\cos b} > 1$ to $\frac 1{\sin A\cos B} > 2$. Maybe... I'll use AM-GM and see what happens.

So we need to use AM-GM to prove $\sin A \cos B < \frac 12$ and $\frac 1{\sin A \cos A}> 2$.

Now AM- GM states for $M,N > 0$ thatn $\frac {M+N}2 \ge \sqrt{MN}$ so $\sin A\cos A = \sqrt{\sin^2 A\cos^2 A} \le \frac {\sin^2 A + \cos^2 A}2 = \frac 12$.

ANd that's that.


Another way $\cos A = \sin (90 - A)$. Let $A = 45 + b$ then

$\sin A = \sin (45+b) = \sin 45 \cos b + \cos 45 \sin b=\frac {\sqrt 2}2(\cos b + \sin b)$ and $\cos A = \sin (45 - b) = \sin 45 \cos b - \cos 45 \sin b=\frac {\sqrt 2}2(\cos b - \sin b)$.

And $\sin A*\cos A = (\frac {\sqrt 2}2)^2(\cos b + \sin b)(\cos b - \sin b) = \frac 12(\cos^2 b - \sin^2 b)= \frac 12(1-2\sin^2 b) < \frac 12$

So $(1 +\frac 1{\sin A})(1 + \frac 1 {\cos A}) =$

$1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1 {\sin A\cos B} > 1 + 1+1+2 =5$.

  • 1
    $\begingroup$ What is $B$, and why can you write $\sin^2 A + \cos^2 B = 1$? $\endgroup$ – aschepler Jul 3 '18 at 0:30
  • $\begingroup$ B is a typo for A. $\endgroup$ – fleablood Jul 3 '18 at 5:10

Let $y=\dfrac{1+\sin A+\cos A+\sin A\cos A}{\sin A\cos A}$

$\iff(y-1)\sin A\cos A-1=\sin A+\cos A$

Squaring both sides we get $$1+\sin2A=1+\dfrac{(y-1)^2\sin^22A}4-(y-1)\sin2A$$


As $\sin2A>0,$ $$(y-1)^2\sin2A-4y=0$$


$\iff y^2-6y+1\ge0\implies y>\dfrac{6+\sqrt{32}}2=3+2\sqrt2$ as $y>0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.